# Tag Info

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Thare is only one node connecting elements B, D, E, and G. The dots on the diagram are just to indicate that the lines do in fact connect, so that all of the wires are part of the same node. In this kind of schematic diagram wires are considered as ideal, and all points on the wire are considered to be at an equal potential.

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Earthing something means dumping the electron flow into the earth. Since the earth is so big, it can absorbe/give a practically infinite amount of charge without changing potential, this means that you can treat earth as a reservoir of ready to use electrons. If you plug the phase of your home power line into the ground (without safety devices in the ...

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Think of it in terms of current, V, W, and Z are in series, so each is equally bright. X and Y are in parallel, so each gets half the current of the others. If you assume each bulb is a constant resistance R (not true for incandescent bulbs, by the way), then V,W and Z will each dissipate $i^2R$. For X and Y, since each has a current i/2, the power will be ...

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There are basically two kinds of addition here (ordinary addition and inverse-sum-inverse), representing series and parallel arrangements. You can represent the thing as a tree with alternating nodes of addition and ISI layers. The thing resolves pretty much down to a tree with N leaves. The magic is dealt with here http://oeis.org/A000669 . It talks ...

2

The Doppler shift for small speeds is $\Delta f/f = \Delta v/c$, where $\Delta v$ is the (signed) speed of the source relative to the detector, and I'm using $c$ as the speed of sound. So let's plug in some numbers. I'm going to use numbers that will produce a large effect to see how larger an effect is plausible. Let's take a woofer operating at $f = 200 ... 2 If you want to include "all real world effects" in your analysis, you need to make sure you include all effects. At the very least, include parasitics. And include the fact that your "real world voltage source" has finite impedance, output capacitance, inductance in the leads, ... So when you state Say it starts of at a voltage V when you connect it to ... 1 The voltage across$R_1$and$R_2$will be the same,$V$. You are right that the net resistance will decrease to$(1/R_1+1/R_2)^{-1}$and this change will be compensated by an increase in the current$I$(Ohm's Law).$V$will stay constant. Why? Because each component in a parallel circuit has two common nodes with each of the other components in the ... 1 The conceptual problem here is that of EMF,$\mathcal{E}$vs Electric Potential, V. They aren't really the same thing despite being measured in the same units. For instance the EMF is caused by an external agent that isn't the conservative electrostatic field, like say a chemical reaction in a battery or a solar cell. Work is done to cause a charge ... 1 Electrochemical cells (batteries) are not passive components, instead they're active charge-pumps having internal feedback effects which produces a relatively constant voltage at the output terminals. If an external field impinges on a battery's terminals, this will produce a temporary small change in potential on the terminals. But the battery then ... 1 See here. There's instructions on how to calcuate$V_{th}$and$R_{th}$. Let$I_3$be the current through$R_3$,$I_1$through$R_1$, etc. With$V_{AB}$open, by KVL, we have$V_1 - R_1I_1 - R_3 I_3 = 0$and$V_1 - R_1 I_1 - R_2 I_2 -V_{AB} = 0$, but when$V_{AB}$is open we have$I_2 = 0$and so$I_1 = I_3$, so$V_1 = (R_1 + R_3) ...

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Both examples are using the same methods. In the first example, the break in the circuit removes the 6 ohm and 5 ohm resistors altogether. All of the current then flows through the 2 ohm resistor, causing the Thevenin resistance to be 2 ohms. In the second example, the current first passes through $R_2$, then is split between $R_1$ and $R_3$ in parallel. ...

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The two currents $i$ and $i_1$ could have been chosen to have whichever direction the author wanted. The choice is arbitrary. The equation $i+ i_1=0$ is just what KCL tells us: the net current leaving any node in the circuit is 0. Since $i$ and $i_1$ are both defined to be leaving the node that connects them, their sum must be zero. This means that one of ...

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Perhaps you meant intuitively? Well, if you imagine the roads are tracks and the flow of cars the current, and if you had the choice between a very bumpy narrow path on your left and a very nice motorway on your right, which would you choose? Nearly everyone would go right (there are some dirt enthusiastic or people that don't think), and overall the current ...

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